Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 No Rob. Cantor's number would be .11111... (done instantly) while your list would never contain .1111... The reason your list does not contain .111... is that it would have to be the last row of your list, but there is no last row. I tried to explain that earlier. Analytics tried to bring it to your attention again. There is no infinitieth row (no last row). That's why we know your list doesn't contain .1111... even though Cantor's number derived from your list would be precisely .1111...Anyhow, I think we're back to our infinite loop.So, let me get this staright. You are suggestingt hat my list could really exist and that indeed a new number could be made from it but that whereas my list is only of finite length, the new number made from that finite number would be infinite? Thats a contradiction. The new number is precisely made from my list. Logic would state that any new number made would be of the same size or "1" bigger than the last number in my list. Thats all fine and dandy but it only shows one thing- no list could really be of infinite length just as no new number could really be infinitly long.Oh, BTW, my list showing that pattern of .11111... is also done "instantly" just as his new number is "done instantly". You couldn't prove to me that my list couldn't be done instantly either just because it shows a pattern.I await.
Zeta-Flux Posted January 30, 2012 Posted January 30, 2012 Now of course we both know that nothing really counts to an actual infinity.If you accept that as a given, there is no point in the argument, because you cannot even make your infinite list from which to find a number not on the list.The set of natural numbers is not a set, because it truly is infinite.----------------------Let me ask a question. Since you don't think there really is anything infinite, what is the largest number?
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 (It's at this time that special rules must be made for Cantor's argument to be true. Of course, all these special rules follow no real mathematical discipline or true principle of number theory)
asbestosman Posted January 30, 2012 Posted January 30, 2012 No, your list is infinite but it does not contain the number .1111... Each number on your list will end in an infinite stream of 0's after the 1's. There are an endless supply of such numbers just as there are an endless supply of powers of one tenth (1/10, 1/100, 1/1000, ... = .100000..., .01000..., .001..., ...) all of which have an endless supply of 0's after the 1.
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 If you accept that as a given, there is no point in the argument, because you cannot even make your infinite list from which to find a number not on the list.The set of natural numbers is not a set, because it truly is infinite.----------------------Let me ask a question. Since you don't think there really is anything infinite, what is the largest number?It wouldn't be any "one" number as one more could always be added to it forever and ever. At the same time, the principle agreeing with the rule, a number is only a number when it is of finite length or has a finite or "known" limit. There is no such thing as an infinitley long number or set of things.
asbestosman Posted January 30, 2012 Posted January 30, 2012 (It's at this time that special rules must be made for Cantor's argument to be true. Of course, all these special rules follow no real mathematical discipline or true principle of number theory)We call these special rules "axioms". The particular axiom in question is the Axiom of Infinity. That one basically says that it makes sense to speak of the set of natural numbers as a set.
asbestosman Posted January 30, 2012 Posted January 30, 2012 It's true, Rob, that all integers are finite. Despite this, there are an endless supply of integers (they do not end). Do you agree?
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 No, your list is infinite but it does not contain the number .1111... Each number on your list will end in an infinite stream of 0's after the 1's. There are an endless supply of such numbers just as there are an endless supply of powers of one tenth (1/10, 1/100, 1/1000, ... = .100000..., .01000..., .001..., ...) all of which have an endless supply of 0's after the 1.Would you agree that there could be no possible end to the "1's" in my list? For if there was, then we would have a contradiction.
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 It's true, Rob, that all integers are finite. Despite this, there are an endless supply of integers (they do not end). Do you agree?I agree
asbestosman Posted January 30, 2012 Posted January 30, 2012 (edited) Would you agree that there could be no possible end to the "1's" in my list? For if there was, then we would have a contradiction.It depends on what you mean. Be careful. There is an endless supply of rows in your list. Each row will have a finite number of 1's in it. There is no end the number of rows, but any individual row does have an end to the number of 1's in it.That may sound like a contradiction, but it is not. That's basically the same thing as the next statement about integers. Each integer is finite, but there are an endless supply of them (they do not end). Edited January 30, 2012 by asbestosman
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 It depends on what you mean. Be careful. There is an endless supply of rows in your list. Each row will have a finite number of 1's in it. There is no end the number of rows, but any individual row does have an end to the number of rows in it.That may sound like a contradiction, but it is not. That's basically the same thing as the next statement about integers. Each integer is finite, but there are an endless supply of them (they do not end).But the argument is that I have an infinite list of rows and that in each row there is a number. If there are truly an infinite amount of rows, then it holds true also that the string of 1's would also get to be of infinite length.This is where I certainly must question Cantor because of the ultimate impossibility that something could really be gone through that was infinitely long.
asbestosman Posted January 30, 2012 Posted January 30, 2012 No. Each row has as many 1's as the integer value for the row. Since all integers are finite, the number of 1's in the row is finite.
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 No. Each row has as many 1's as the integer value for the row. Since all integers are finite, the number of 1's in the row is finite.So then, if Cantor were to build a new number from my list it too would be finite.
asbestosman Posted January 30, 2012 Posted January 30, 2012 So then, if Cantor were to build a new number from my list it too would be finite.It would be one finite decimal number consisting of an endless supply of 1's.Each row on your list has a finite number of 1's, but an endless supply of 0's thereafter.
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 Who would have more "1's" in their number- one of the numbers on my list or the new number Cantor is making?
asbestosman Posted January 30, 2012 Posted January 30, 2012 The new number Cantor is making. Each number on your list has a finite number of 1's. The new number Cantor is making has an endless supply of 1's.
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 It would be one finite decimal number consisting of an endless supply of 1's.Each row on your list has a finite number of 1's, but an endless supply of 0's thereafter.But it's supposedly done all at once, hence the contradiction. My list has the capacity to supply an endless supply of 1's for each new row which themselves are endless in supply. Cantor's number is only constructed from my list which must begin at the start and continue row by row and be counted as a natural number is counted for each operation.
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 The new number Cantor is making. Each number on your list has a finite number of 1's. The new number Cantor is making has an endless supply of 1's.But Cantor is making a new number from my list of numbers. If he truly has more 1's than any number in my list, it could only be "1" more which is a contradiction about the notion of the argument anyway that it could end.
asbestosman Posted January 30, 2012 Posted January 30, 2012 If you took the first column from your list starting on the second row, that number would be .1111... the same as Cantor's. You have an endless supply of 1's that direction because there are an endless supply of integers even though each integer is finite. You do have an endless supply of 1's, going down because you have an endless supply of rows.butYou do not have any row with an endless supply of 1's in that row for the same reason that every integer is finite.
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 So, mathematically speaking, would his new number be best represented like this- .11111...
asbestosman Posted January 30, 2012 Posted January 30, 2012 But Cantor is making a new number from my list of numbers. If he truly has more 1's than any number in my list, it could only be "1" more which is a contradiction about the notion of the argument anyway that it could end.No, he has infinitely more. He makes it from your list, but your list has an endless supply of 0's on the diagonal. Each row only has a finite supply of 1's just as each integer is a finite number although there are an endless supply of integers.
asbestosman Posted January 30, 2012 Posted January 30, 2012 So, mathematically speaking, would his new number be best represented like this- .11111...That works. I might also call it one ninth.
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 That works. I might also call it one ninth.So bear with me for a second then. If his number is best represented as .1111... this would mean that his list was actually constructed from a list with the previous row having also an infinite count of 1's in it otherwise his new number couldn't be represented like this. We are going to use another proof to show just how many 1's are in the last row or in the new number. Her eit is- We use the same set only this time we start with it like this-.1.11.111.1111.11111.111111etc.Now we are going to make a number only this time we are going to construct it in this manner- our first number is the same as found in the first row, etc. Basically- we still move in diagonalization only we copy the same number as is listed for that placement. So, our number will be .111111... Logically, the number created would thus be best represented as ".1111..." but, it would be equal in length as the last row it was ever to count. Thus if the new number is ".1111..." then we know it is equal to the last row in the list which would also be .1111...
asbestosman Posted January 30, 2012 Posted January 30, 2012 So bear with me for a second then. If his number is best represented as .1111... this would mean that his list was actually constructed from a list with the previous row having also an infinite count of 1's in it otherwise his new number couldn't be represented like this.I don't know what you mean here. What do you mean by a previous row having an infinite count of 1's in it? Previous to what row? If the previous row has an infinite count of 1's in it, how many are in the row itself? 0? infinity+1? I'm lost here.We are going to use another proof to show just how many 1's are in the last row or in the new number. Her eit is- We use the same set only this time we start with it like this-.1.11.111.1111.11111.111111etc.Now we are going to make a number only this time we are going to construct it in this manner- our first number is the same as found in the first row, etc. Basically- we still move in diagonalization only we copy the same number as is listed for that placement. So, our number will be .111111... Logically, the number created would thus be best represented as ".1111..." but, it would be equal in length as the last row it was ever to count. Thus if the new number is ".1111..." then we know it is equal to the last row in the list which would also be .1111...Only one problem with your proof. Repeat with me. Slowly. There is no last row in Rob's list. There is no infinitieth row in Rob's list.
Rob Osborn Posted January 30, 2012 Author Posted January 30, 2012 I don't know what you mean here. What do you mean by a previous row having an infinite count of 1's in it? Previous to what row? If the previous row has an infinite count of 1's in it, how many are in the row itself? 0? infinity+1? I'm lost here.The new number, which you agreed with is best represented as ".1111..." Beacuse the new number can only be constructed with the same amount of rows, then there must be infinite rows that were counted in the number. That being the case, the last row that was counted to get the new number- ".1111..." also had a count of ".1111..." in it otherwise the proof is wrong.Only one problem with your proof. Repeat with me. Slowly. There is no last row in Rob's list. There is no infinitieth row in Rob's list.Then there can be no new number that is represented as ".1111..." You can't have it both ways. Get the drift? If the new number really is ".1111..." as you admit then one must also admit that it has counted an infinity of rows.So, it;s either there are an infinity amount of rows and thus a new number ".1111..." or the new number is only a finite number and can't be represented as ".1111..." The last proof I gave shows that this principle is correct. If we are to give the diagonal count expressed as "n" and the count of rows as "x" then it stands to reason that n=x, they have the same cardinality. This thus proves that whatever we make as "n" must always have a one to one correspondence with "x"
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